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As stated in the title, it is requested to define a linear transformation $T:\Bbb R^3 \to \Bbb R^3$ such that the null space of $T$ is the $z$-axis, and the range of $T$ is the plane: $x+y+z=0$


I don't really know how to begin with the solution of the exercise, I think that I should try to get a matrix using the standard base, but after that, I don't have any concrete ideas.

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You have to find a basis of that plane: $x+y+z=0$ then $x=-y-z$ so you can pick $v_1=(1,-1,0), v_2=(1,0,-1)$. The $z$-axis is the vector $e_3=(0,0,1)$.

If you define your map in a basis you are done. In particular take the standard basis, then

$f(e_1)=v_1, f(e_2)=v_2, f(e_3)=(0,0,0)$

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  • $\begingroup$ I still have one doubt, for the particular choice for the vectors. we know it is a plane, therefore its dimension is 2, then all we need is to find two linearly independent vectors that satisfy the condition that the range is inside the plane. Am I right? $\endgroup$ – zastenAmir Jun 1 at 7:04
  • $\begingroup$ Exactly! You were right! A basis of that plane consits of two vectors (those ones i wrote in my answer). Moreover, since you know the kernel has dimension one, then by rank-null formula, you know that the range has dimension 2 $\endgroup$ – Davide Motta Jun 1 at 7:43
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Choose a basis $\mathcal{B}=\{u,v\}$ for $x+y+z=0$ (for instance, it could be $\{(2,-1,-1), (1,1,-2)\}$).

Letting $T$ be the linear extension of $$\left\{\begin{array}{c} e_1\mapsto u \\ e_2 \mapsto v \\ e_3\mapsto 0\end{array}\right.$$ should work.

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